All categories

3 years ago

A decorator wants to line the bottom of 3 drawers with paper. If the bottom of each drawer measures 36 inches by 20 inches,

Mathematics

2 answers:

prisoha [69]3 years ago

6 0

Answer:

2160

Step-by-step explanation:

36*20=720 720*3=2160

kvasek [131]3 years ago

3 0

Answer: D 2,160

Step-by-step explanation:

Given : The bottom of each drawer measures 36 inches by 20 inches.

Then area of bottom of each drawer is given :-

$\text{Area}=36\times20=720\ \text{ square inches}$

Now, the area for 3 drawers will be :-

$\text{Area}=3\times720=2160\text{ square inches}$

Hence, the area of paper needed = 2160 square inches

You might be interested in

Given that the circumferential of a circle is 22cm calculate its diameter

Ipatiy [6.2K]

Answer:

diameter =7 cm

Step-by-step explanation:

circum=2pi×r

$2\pi \times r$

2×pi×r=22

2×22÷7×r=22

r=22×7÷44

r=7÷2

r=3.5

d=2×r=2×3.5=7

6 0

2 years ago

18 packs of socks fit into a box.

sukhopar [10]

Answer:

no. of packs in one box = 18

no. of boxes = 7

total no. of packs of socks = 18*7

=126

7 0

3 years ago

Read 2 more answers

What’s the area of a circle with a radius of 1 meter. Express in terms of pie

Anni [7]

Answer:

A=pi

Step-by-step explanation:

The formula of the area of a circle is A=pi*r^2 where r=radius.

In this problem, you have radius of 1 meter. So just plug that into the equation and solve for the area.

A=pi*(1)^2=pi*1*1=pi*1=pi

A=pi

8 0

3 years ago

What is the number of possible outcomes for the tree diagram below?

pav-90 [236]

Answer:

i think the answer is ( C )

8 0

3 years ago

Help me solve them plsss...both of the questions :)

zaharov [31]

$\bold{\text{Answer:}\quad \dfrac{\sqrt2}{144}}$

<u>Step-by-step explanation:</u>

When you plug in 3 directly to the equation, you get 0/0

Since it is indeterminate, you have to use L'Hopital's Rule.

That means that you find the limit of the DERIVATIVE of the numerator and the DERIVATIVE of the denominator.

$\dfrac{d}{dx}(\sqrt{5+\sqrt{2x+3}}-2\sqrt2)\quad = \dfrac{1}{2\sqrt{5+\sqrt{2x+3}}(\sqrt{2x+3})}\\\\\\f'(3)\ \text{for the numerator}\ =\dfrac{1}{12\sqrt2}\\\\\\\\\dfrac{d}{dx}(x^2-9) = 2x\\\\\\f'(3) \text{for the denominator}\quad = 6\\\\\\\dfrac{f'(3)\ \text{numerator}}{f'(3)\ \text{denominator}}\quad = \dfrac{\dfrac{1}{12\sqrt2}}{6}\quad = \dfrac{1}{72\sqrt2}\quad \rightarrow \large\boxed{\dfrac{\sqrt2}{144}}$

4 0

3 years ago

Read 2 more answers